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Molecular symmetry

In this chapter we sharpen the concept of ‘shape’ into a precise definition of ‘symmetry’,and show that symmetry may be discussed systematically. We see how to classify anymolecule according to its symmetry and how to use this classification to discuss molecularproperties. After describing the symmetry properties of molecules themselves, we turn to a consideration of the effect of symmetry transformations on orbitals and see that theirtransformation properties can be used to set up a labelling scheme. These symmetry labelsare used to identify integrals that necessarily vanish. One important integral is the overlap integral between two orbitals. By knowing which atomic orbitals may have nonzero overlap,we can decide which ones can contribute to molecular orbitals. We also see how to selectlinear combinations of atomic orbitals that match the symmetry of the nuclear framework.Finally, by considering the symmetry properties of integrals, we see that it is possible to derive the selection rules that govern spectroscopic transitions.

The systematic discussion of symmetry is called group theory. Much of group theoryis a summary of common sense about the symmetries of objects. However, becausegroup theory is systematic, its rules can be applied in a straightforward, mechanicalway. In most cases the theory gives a simple, direct method for arriving at useful con-clusions with the minimum of calculation, and this is the aspect we stress here. Insome cases, though, it leads to unexpected results.

The symmetry elements of objects

Some objects are ‘more symmetrical’ than others. A sphere is more symmetrical thana cube because it looks the same after it has been rotated through any angle about anydiameter. A cube looks the same only if it is rotated through certain angles aboutspecific axes, such as 90°, 180°, or 270° about an axis passing through the centres ofany of its opposite faces (Fig. 12.1), or by 120° or 240° about an axis passing throughany of its opposite corners. Similarly, an NH3 molecule is ‘more symmetrical’ than anH2O molecule because NH3 looks the same after rotations of 120° or 240° about theaxis shown in Fig. 12.2, whereas H2O looks the same only after a rotation of 180°.

An action that leaves an object looking the same after it has been carried out is calleda symmetry operation. Typical symmetry operations include rotations, reflections,and inversions. There is a corresponding symmetry element for each symmetry opera-tion, which is the point, line, or plane with respect to which the symmetry operationis performed. For instance, a rotation (a symmetry operation) is carried out around anaxis (the corresponding symmetry element). We shall see that we can classify moleculesby identifying all their symmetry elements, and grouping together molecules that

The symmetry elements ofobjects

12.1 Operations and symmetryelements

12.2 The symmetry classification ofmolecules

12.3 Some immediateconsequences of symmetry

Applications to molecularorbital theory and spectroscopy

12.4 Character tables and symmetrylabels

12.5 Vanishing integrals and orbitaloverlap

12.6 Vanishing integrals andselection rules

Checklist of key ideas

Further reading

Discussion questions

Exercises

Problems

12

12.1 OPERATIONS AND SYMMETRY ELEMENTS 405

possess the same set of symmetry elements. This procedure, for example, puts the trigonal pyramidal species NH3 and SO3

2− into one group and the angular species H2Oand SO2 into another group.

12.1 Operations and symmetry elements

The classification of objects according to symmetry elements corresponding to opera-tions that leave at least one common point unchanged gives rise to the point groups.There are five kinds of symmetry operation (and five kinds of symmetry element) ofthis kind. When we consider crystals (Chapter 20), we shall meet symmetries arisingfrom translation through space. These more extensive groups are called space groups.

The identity, E, consists of doing nothing; the corresponding symmetry element isthe entire object. Because every molecule is indistinguishable from itself if nothing isdone to it, every object possesses at least the identity element. One reason for includ-ing the identity is that some molecules have only this symmetry element (1); anotherreason is technical and connected with the detailed formulation of group theory.

An n-fold rotation (the operation) about an n-fold axis of symmetry, Cn (the cor-responding element) is a rotation through 360°/n. The operation C1 is a rotationthrough 360°, and is equivalent to the identity operation E. An H2O molecule has onetwofold axis, C2. An NH3 molecule has one threefold axis, C3, with which is associatedtwo symmetry operations, one being 120° rotation in a clockwise sense and the other120° rotation in a counter-clockwise sense. A pentagon has a C5 axis, with two (clock-wise and counterclockwise) rotations through 72° associated with it. It also has an axisdenoted C5

2, corresponding to two successive C5 rotations; there are two such opera-tions, one through 144° in a clockwise sense and the other through 144° in a counter-clockwise sense. A cube has three C4 axes, four C3 axes, and six C2 axes. However, eventhis high symmetry is exceeded by a sphere, which possesses an infinite number ofsymmetry axes (along any diameter) of all possible integral values of n. If a moleculepossesses several rotation axes, then the one (or more) with the greatest value of n iscalled the principal axis. The principal axis of a benzene molecule is the sixfold axisperpendicular to the hexagonal ring (2).

C2

C3

C4

Fig. 12.1 Some of the symmetry elements ofa cube. The twofold, threefold, andfourfold axes are labelled with theconventional symbols.

C2

C3

NH

O

H

(a)

(b)

Fig. 12.2 (a) An NH3 molecule has athreefold (C3) axis and (b) an H2Omolecule has a twofold (C2) axis. Both have other symmetry elements too.

F

Cl

Br

IC

1 CBrClFI

C6

2 Benzene, C H6 6

Comment 12.1

There is only one twofold rotationassociated with a C2 axis becauseclockwise and counter-clockwise 180°rotations are identical.

406 12 MOLECULAR SYMMETRY

A reflection (the operation) in a mirror plane, σ (the element), may contain theprincipal axis of a molecule or be perpendicular to it. If the plane is parallel to theprincipal axis, it is called ‘vertical’ and denoted σv. An H2O molecule has two verticalplanes of symmetry (Fig. 12.3) and an NH3 molecule has three. A vertical mirror planethat bisects the angle between two C2 axes is called a ‘dihedral plane’ and is denoted σd (Fig. 12.4). When the plane of symmetry is perpendicular to the principal axis it is called ‘horizontal’ and denoted σh. A C6H6 molecule has a C6 principal axis and ahorizontal mirror plane (as well as several other symmetry elements).

In an inversion (the operation) through a centre of symmetry, i (the element), weimagine taking each point in a molecule, moving it to the centre of the molecule, andthen moving it out the same distance on the other side; that is, the point (x, y, z) istaken into the point (−x, −y, −z). Neither an H2O molecule nor an NH3 molecule hasa centre of inversion, but a sphere and a cube do have one. A C6H6 molecule does havea centre of inversion, as does a regular octahedron (Fig. 12.5); a regular tetrahedronand a CH4 molecule do not.

An n-fold improper rotation (the operation) about an n-fold axis of improperrotation or an n-fold improper rotation axis, Sn, (the symmetry element) is com-posed of two successive transformations. The first component is a rotation through360°/n, and the second is a reflection through a plane perpendicular to the axis of thatrotation; neither operation alone needs to be a symmetry operation. A CH4 moleculehas three S4 axes (Fig. 12.6).

12.2 The symmetry classification of molecules

To classify molecules according to their symmetries, we list their symmetry elementsand collect together molecules with the same list of elements. This procedure puts CH4

and CCl4, which both possess the same symmetry elements as a regular tetrahedron,into the same group, and H2O into another group.

The name of the group to which a molecule belongs is determined by the symmetryelements it possesses. There are two systems of notation (Table 12.1). The Schoenfliessystem (in which a name looks like C4v) is more common for the discussion of indi-vidual molecules, and the Hermann–Mauguin system, or International system (inwhich a name looks like 4mm), is used almost exclusively in the discussion of crystalsymmetry. The identification of a molecule’s point group according to the Schoenfliessystem is simplified by referring to the flow diagram in Fig. 12.7 and the shapes shownin Fig. 12.8.

vv́

s

s

d

dd

s

s

s

Fig. 12.3 An H2O molecule has two mirrorplanes. They are both vertical (i.e. containthe principal axis), so are denoted σv and σv′ .

Fig. 12.4 Dihedral mirror planes (σd) bisectthe C2 axes perpendicular to the principalaxis.

Centre ofinversion,i

Fig. 12.5 A regular octahedron has a centreof inversion (i).

h

h

C4

S4

C6

S6

(a)

(b)

s

s

Fig. 12.6 (a) A CH4 molecule has a fourfoldimproper rotation axis (S4): the molecule is indistinguishable after a 90° rotationfollowed by a reflection across thehorizontal plane, but neither operationalone is a symmetry operation. (b) Thestaggered form of ethane has an S6 axiscomposed of a 60° rotation followed by areflection.

12.2 THE SYMMETRY CLASSIFICATION OF MOLECULES 407

Cnv

S2n

Molecule

Linear?Y N

i?Y N Y

N

Twoor more

, > 2?C nn

Y Ni?

C 5?Y NIh

Td

YY

Y

N

?Cs

Nh?

Y

N

n d?Dnd Dn

Y N

h?Y

N

N

Cnh

Y n v?

S2n?Y N

Cn

N

Y Ni?C i

Select withhighest ; then, are theperpendicular to ?

Cn nC

C

n

n

2

Dnh

Cn?

D¥h C

¥v

Oh

C1

s

s

s

s

s

Fig. 12.7 A flow diagram for determining the point group of a molecule. Start at the top andanswer the question posed in each diamond (Y = yes, N = no).

(a) The groups C1, Ci, and Cs

A molecule belongs to the group C1 if it has no element other than the identity, as in(1). It belongs to Ci if it has the identity and the inversion alone (3), and to Cs if it hasthe identity and a mirror plane alone (4).

COOHOHH

COOH

HOH

Centre ofinversion

3 Meso-tartaric acid,HOOCCH(OH)CH(OH)COOH

N

4 Quinoline, C H N9 7


Cone

2 3 4 5 6 ¥

Cn

Dn

Cnv

Dnh

Dnd

Cnh

S2n

(plane or bipyramid)

(pyramid)

n =

Fig. 12.8 A summary of the shapes corresponding to different point groups. The group towhich a molecule belongs can often be identified from this diagram without going throughthe formal procedure in Fig. 12.7.

Table 12.1 The notation for point groups*

Ci ⁄

Cs m

C1 1 C2 2 C3 3 C4 4 C6 6

C2v 2mm C3v 3m C4v 4mm C6v 6mm

C2h 2m C3h fl C4h 4/m C6h 6/m

D2 222 D3 32 D4 422 D6 622

D2h mmm D3h fl2m D4h 4/mmm D6h 6/mmm

D2d ›2m D3d ‹m S4 › /m S6 3–

T 23 Td ›3m Th m3

O 432 Oh m3m

* In the International system (or Hermann–Mauguin system) for point groups, a number n denotes thepresence of an n-fold axis and m denotes a mirror plane. A slash (/) indicates that the mirror plane isperpendicular to the symmetry axis. It is important to distinguish symmetry elements of the same type but ofdifferent classes, as in 4/mmm, in which there are three classes of mirror plane. A bar over a number indicatesthat the element is combined with an inversion. The only groups listed here are the so-called ‘crystallographicpoint groups’ (Section 20.1).

12.2 THE SYMMETRY CLASSIFICATION OF MOLECULES 409

(b) The groups Cn, Cnv, and Cnh

A molecule belongs to the group Cn if it possesses an n-fold axis. Note that symbol Cn

is now playing a triple role: as the label of a symmetry element, a symmetry operation,and a group. For example, an H2O2 molecule has the elements E and C2 (5), so it belongs to the group C2.

If in addition to the identity and a Cn axis a molecule has n vertical mirror planes σv,then it belongs to the group Cnv. An H2O molecule, for example, has the symmetryelements E, C2, and 2σv, so it belongs to the group C2v. An NH3 molecule has the ele-ments E, C3, and 3σv, so it belongs to the group C3v. A heteronuclear diatomicmolecule such as HCl belongs to the group C∞v because all rotations around the axisand reflections across the axis are symmetry operations. Other members of the groupC∞v include the linear OCS molecule and a cone.

Objects that in addition to the identity and an n-fold principal axis also have a hor-izontal mirror plane σh belong to the groups Cnh. An example is trans-CHCl=CHCl(6), which has the elements E, C2, and σh, so belongs to the group C2h; the moleculeB(OH)3 in the conformation shown in (7) belongs to the group C3h. The presence ofcertain symmetry elements may be implied by the presence of others: thus, in C2h theoperations C2 and σh jointly imply the presence of a centre of inversion (Fig. 12.9).

(c) The groups Dn, Dnh, and Dnd

We see from Fig. 12.7 that a molecule that has an n-fold principal axis and n twofoldaxes perpendicular to Cn belongs to the group Dn. A molecule belongs to Dnh if it alsopossesses a horizontal mirror plane. The planar trigonal BF3 molecule has the ele-ments E, C3, 3C2, and σh (with one C2 axis along each B-F bond), so belongs to D3h

(8). The C6H6 molecule has the elements E, C6, 3C2, 3C2′, and σh together with someothers that these elements imply, so it belongs to D6h. All homonuclear diatomicmolecules, such as N2, belong to the group D∞h because all rotations around the axisare symmetry operations, as are end-to-end rotation and end-to-end reflection; D∞h

is also the group of the linear OCO and HCCH molecules and of a uniform cylinder.Other examples of Dnh molecules are shown in (9), (10), and (11).

A molecule belongs to the group Dnd if in addition to the elements of Dn it possessesn dihedral mirror planes σd. The twisted, 90° allene (12) belongs to D2d, and the stag-gered conformation of ethane (13) belongs to D3d.

(d) The groups Sn

Molecules that have not been classified into one of the groups mentioned so far, butthat possess one Sn axis, belong to the group Sn. An example is tetraphenylmethane,which belongs to the point group S4 (14). Molecules belonging to Sn with n > 4 arerare. Note that the group S2 is the same as Ci, so such a molecule will already have beenclassified as Ci.

O

H

C2

5 Hydrogen peroxide, H O2 2

h

C2

is

Fig. 12.9 The presence of a twofold axis anda horizontal mirror plane jointly imply thepresence of a centre of inversion in themolecule.

Cl

Cl

C2

6 CHCl=CHCltrans-

h$

B

OH

7 B(OH)3

C3

h$

B

F

8 Boron trifluoride, BF3

Comment 12.2

The prime on 3C2′ indicates that thethree C2 axes are different from the otherthree C2 axes. In benzene, three of the C2axes bisect C-C bonds and the otherthree pass through vertices of thehexagon formed by the carbonframework of the molecule.


(e) The cubic groups

A number of very important molecules (e.g. CH4 and SF6) possess more than oneprincipal axis. Most belong to the cubic groups, and in particular to the tetrahedralgroups T, Td, and Th (Fig. 12.10a) or to the octahedral groups O and Oh (Fig. 12.10b).A few icosahedral (20-faced) molecules belonging to the icosahedral group, I(Fig. 12.10c), are also known: they include some of the boranes and buckminster-fullerene, C60 (15). The groups Td and Oh are the groups of the regular tetrahedron(for instance, CH4) and the regular octahedron (for instance, SF6), respectively. If the object possesses the rotational symmetry of the tetrahedron or the octahedron,but none of their planes of reflection, then it belongs to the simpler groups T or O(Fig. 12.11). The group Th is based on T but also contains a centre of inversion (Fig. 12.12).

P

Cl

C3

C2

C2

C2

h

10 Phosphorus pentachloride,PCl ( )5 3hD

$

h

C4

C2

C2 C2

–Cl

Au

11 Tetrachloroaurate(III) ion,[AuCl ] ( )4 4h

� D

$

h

C2

C2

9 Ethene, CH =CH ( )2 2 2hD

$

C2C2

C S2 4,

12 Allene, C H ( )3 4 2dD

d

C2C S3 3,

13 Ethane, C H ( )2 6 3dD

$

S4

Ph

14 Tetraphenylmethane,C(C H ) ( )6 5 4 4S

(a) (b) (c)

15 Buckminsterfullerene, C ( )60 I

Fig. 12.10 (a) Tetrahedral, (b) octahedral,and (c) icosahedral molecules are drawn ina way that shows their relation to a cube:they belong to the cubicgroups Td, Oh, andIh, respectively.

12.3 SOME IMMEDIATE CONSEQUENCES OF SYMMETRY 411

(a) T (b) O

Fig. 12.11 Shapes corresponding to the point groups (a) T and (b) O. The presence of thedecorated slabs reduces the symmetry of the object from Td and Oh, respectively.

Fig. 12.12 The shape of an object belongingto the group Th.

cp (= C H )5 5

Ru

16 Ruthenocene, Ru(cp)2

Comment 12.3

The web site contains links to interactivetutorials, where you use your webbrowser to to assign point groups ofmolecules.

cp (= C H )5 5

Fe

17 Ferrocene, Fe(cp)2

(f ) The full rotation group

The full rotation group, R3 (the 3 refers to rotation in three dimensions), consists of an infinite number of rotation axes with all possible values of n. A sphere and anatom belong to R3, but no molecule does. Exploring the consequences of R3 is a very important way of applying symmetry arguments to atoms, and is an alternative approach to the theory of orbital angular momentum.

Example 12.1 Identifying a point group of a molecule

Identify the point group to which a ruthenocene molecule (16) belongs.

Method Use the flow diagram in Fig. 12.7.

Answer The path to trace through the flow diagram in Fig. 12.7 is shown by a greenline; it ends at Dnh. Because the molecule has a fivefold axis, it belongs to the groupD5h. If the rings were staggered, as they are in an excited state of ferrocene that lies4 kJ mol−1 above the ground state (17), the horizontal reflection plane would beabsent, but dihedral planes would be present.

Self-test 12.1 Classify the pentagonal antiprismatic excited state of ferrocene (17).[D5d]

12.3 Some immediate consequences of symmetry

Some statements about the properties of a molecule can be made as soon as its pointgroup has been identified.

(a) Polarity

A polar molecule is one with a permanent electric dipole moment (HCl, O3, and NH3

are examples). If the molecule belongs to the group Cn with n > 1, it cannot possess acharge distribution with a dipole moment perpendicular to the symmetry axis becausethe symmetry of the molecule implies that any dipole that exists in one direction per-pendicular to the axis is cancelled by an opposing dipole (Fig. 12.13a). For example,the perpendicular component of the dipole associated with one O-H bond in H2O iscancelled by an equal but opposite component of the dipole of the second O-H bond,so any dipole that the molecule has must be parallel to the twofold symmetry axis.


However, as the group makes no reference to operations relating the two ends of the molecule, a charge distribution may exist that results in a dipole along the axis (Fig. 12.13b), and H2O has a dipole moment parallel to its twofold symmetry axis. The same remarks apply generally to the group Cnv, so molecules belonging to any of the Cnv groups may be polar. In all the other groups, such as C3h, D, etc., there are symmetry operations that take one end of the molecule into the other. Therefore, as well as having no dipole perpendicular to the axis, such molecules can have nonealong the axis, for otherwise these additional operations would not be symmetry operations. We can conclude that only molecules belonging to the groups Cn , Cnv , andCs may have a permanent electric dipole moment.

For Cn and Cnv, that dipole moment must lie along the symmetry axis. Thus ozone,O3, which is angular and belongs to the group C2v, may be polar (and is), but carbondioxide, CO2, which is linear and belongs to the group D∞h, is not.

(b) Chirality

A chiral molecule (from the Greek word for ‘hand’) is a molecule that cannot be superimposed on its mirror image. An achiral molecule is a molecule that can be superimposed on its mirror image. Chiral molecules are optically active in the sensethat they rotate the plane of polarized light (a property discussed in more detail inAppendix 3). A chiral molecule and its mirror-image partner constitute an enan-tiomeric pair of isomers and rotate the plane of polarization in equal but opposite directions.

A molecule may be chiral, and therefore optically active, only if it does not possess an axis of improper rotation, Sn. However, we need to be aware that such an axis maybe present under a different name, and be implied by other symmetry elements thatare present. For example, molecules belonging to the groups Cnh possess an Sn axisimplicitly because they possess both Cn and σh, which are the two components of an improper rotation axis. Any molecule containing a centre of inversion, i, also pos-sesses an S2 axis, because i is equivalent to C2 in conjunction with σh, and that com-bination of elements is S2 (Fig. 12.14). It follows that all molecules with centres of inversion are achiral and hence optically inactive. Similarly, because S1 = σ, it followsthat any molecule with a mirror plane is achiral.

A molecule may be chiral if it does not have a centre of inversion or a mirror plane,which is the case with the amino acid alanine (18), but not with glycine (19). How-ever, a molecule may be achiral even though it does not have a centre of inversion. Forexample, the S4 species (20) is achiral and optically inactive: though it lacks i (that is,S2) it does have an S4 axis.

(a) (b)

Fig. 12.13 (a) A molecule with a Cn axiscannot have a dipole perpendicular to theaxis, but (b) it may have one parallel to the axis. The arrows represent localcontributions to the overall electric dipole,such as may arise from bonds betweenpairs of neighbouring atoms with differentelectronegativities.

i

S2

Fig. 12.14 Some symmetry elements areimplied by the other symmetry elements ina group. Any molecule containing aninversion also possesses at least an S2

element because i and S2 are equivalent.

COOH

C

NH2

CH3

H

18 -Alanine, NH CH(CH )COOHL 2 3

NH2

C

COOH

H

H

19 Glycine, NH CH COOH2 2

H

N

HH3C

H

H CH3

CH3H3C

S4

20 N(CH CH(CH )CH(CH )CH )2 3 3 2 2

+

+

12.4 CHARACTER TABLES AND SYMMETRY LABELS 413

Applications to molecular orbital theory and spectroscopy

We shall now turn our attention away from the symmetries of molecules themselvesand direct it towards the symmetry characteristics of orbitals that belong to the vari-ous atoms in a molecule. This material will enable us to discuss the formulation andlabelling of molecular orbitals and selection rules in spectroscopy.

12.4 Character tables and symmetry labels

We saw in Chapter 11 that molecular orbitals of diatomic and linear polyatomicmolecules are labelled σ, π, etc. These labels refer to the symmetries of the orbitalswith respect to rotations around the principal symmetry axis of the molecule. Thus, aσ orbital does not change sign under a rotation through any angle, a π orbital changessign when rotated by 180°, and so on (Fig. 12.15). The symmetry classifications σ andπ can also be assigned to individual atomic orbitals in a linear molecule. For example,we can speak of an individual pz orbital as having σ symmetry if the z-axis lies alongthe bond, because pz is cylindrically symmetrical about the bond. This labelling of orbitals according to their behaviour under rotations can be generalized and extendedto nonlinear polyatomic molecules, where there may be reflections and inversions totake into account as well as rotations.

(a) Representations and characters

Labels analogous to σ and π are used to denote the symmetries of orbitals in poly-atomic molecules. These labels look like a, a1, e, eg, and we first encountered them inFig. 11.4 in connection with the molecular orbitals of benzene. As we shall see, theselabels indicate the behaviour of the orbitals under the symmetry operations of the relevant point group of the molecule.

A label is assigned to an orbital by referring to the character table of the group, atable that characterizes the different symmetry types possible in the point group.Thus, to assign the labels σ and π, we use the table shown in the margin. This table isa fragment of the full character table for a linear molecule. The entry +1 shows that theorbital remains the same and the entry −1 shows that the orbital changes sign underthe operation C2 at the head of the column (as illustrated in Fig. 12.15). So, to assignthe label σ or π to a particular orbital, we compare the orbital’s behaviour with theinformation in the character table.

The entries in a complete character table are derived by using the formal techniquesof group theory and are called characters, χ (chi). These numbers characterize the essential features of each symmetry type in a way that we can illustrate by consideringthe C2v molecule SO2 and the valence px orbitals on each atom, which we shall denotepS, pA, and pB (Fig. 12.16).

Under σv, the change (pS, pB, pA) ← (pS, pA, pB) takes place. We can express thistransformation by using matrix multiplication:

(pS, pB, pA) = (pS, pA, pB) = (pS, pA, pB)D(σv) (12.1)

The matrix D(σv) is called a representative of the operation σv. Representatives takedifferent forms according to the basis, the set of orbitals, that has been adopted.

1 0 00 0 10 1 0

⎛

⎝⎜⎜

⎞

⎠⎟⎟

s

p

+

+

-

Fig. 12.15 A rotation through 180° about theinternuclear axis (perpendicular to thepage) leaves the sign of a σ orbitalunchanged but the sign of a π orbital ischanged. In the language introduced in thischapter, the characters of the C2 rotationare +1 and −1 for the σ and π orbitals,respectively.

C2 (i.e. rotation by 180°)

σ +1 (i.e. no change of sign)

π −1 (i.e. change of sign)

Comment 12.4

See Appendix 2 for a summary of therules of matrix algebra.

�

�

�

B�

S �

A �

Fig. 12.16 The three px orbitals that are usedto illustrate the construction of a matrixrepresentation in a C2v molecule (SO2).


We can use the same technique to find matrices that reproduce the other symmetryoperations. For instance, C2 has the effect (−pS, −pB, −pA) ← (pS, pA, pB), and its repres-entative is

D(C2) = (12.2)

The effect of σ ′v is (−pS, −pA, −pB) ← (pS, pA, pB), and its representative is

D(σ ′v) = (12.3)

The identity operation has no effect on the basis, so its representative is the 3 × 3 unitmatrix:

D(E) = (12.4)

The set of matrices that represents all the operations of the group is called a matrixrepresentation, Γ (uppercase gamma), of the group for the particular basis we havechosen. We denote this three-dimensional representation by Γ (3). The discovery of amatrix representation of the group means that we have found a link between symbolicmanipulations of operations and algebraic manipulations of numbers.

The character of an operation in a particular matrix representation is the sum of the diagonal elements of the representative of that operation. Thus, in the basis we areillustrating, the characters of the representatives are

D(E) D(C2) D(σv) D(σ ′v)3 −1 1 −3

The character of an operation depends on the basis.Inspection of the representatives shows that they are all of block-diagonal form:

D =

The block-diagonal form of the representatives show us that the symmetry operationsof C2v never mix pS with the other two functions. Consequently, the basis can be cutinto two parts, one consisting of pS alone and the other of (pA, pB). It is readily verifiedthat the pS orbital itself is a basis for the one-dimensional representation

D(E) = 1 D(C2) = −1 D(σv) = 1 D(σ ′v) = −1

which we shall call Γ (1). The remaining two basis functions are a basis for the two-dimensional representation Γ (2):

D(E) = D(C2) = D(σv) = D(σ v′) =

These matrices are the same as those of the original three-dimensional representa-tion, except for the loss of the first row and column. We say that the original three-dimensional representation has been reduced to the ‘direct sum’ of a one-dimensionalrepresentation ‘spanned’ by pS, and a two-dimensional representation spanned by(pA, pB). This reduction is consistent with the common sense view that the central

−−

⎛⎝⎜

⎞⎠⎟

1 00 1

0 11 0

⎛⎝⎜

⎞⎠⎟

0 11 0

−−

⎛⎝⎜

⎞⎠⎟

1 00 1

⎛⎝⎜

⎞⎠⎟

[ ][ ] [ ][ ] [ ]

n

n n

n n

0 000

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 0 00 1 00 0 1

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−−

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 0 00 1 00 0 1

−−

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 0 00 0 10 1 0


orbital plays a role different from the other two. We denote the reduction symbolicallyby writing

Γ (3) = Γ (1) + Γ (2) (12.5)

The one-dimensional representation Γ (1) cannot be reduced any further, and iscalled an irreducible representation of the group (an ‘irrep’). We can demonstratethat the two-dimensional representation Γ (2) is reducible (for this basis in this group)by switching attention to the linear combinations p1 = pA + pB and p2 = pA − pB. Thesecombinations are sketched in Fig. 12.17. The representatives in the new basis can beconstructed from the old by noting, for example, that under σv, (pB, pA) ← (pA, pB).In this way we find the following representation in the new basis:

D(E) = D(C2) = D(σv) = D(σ v′) =

The new representatives are all in block-diagonal form, and the two combinations arenot mixed with each other by any operation of the group. We have therefore achievedthe reduction of Γ (2) to the sum of two one-dimensional representations. Thus, p1

spans

D(E) = 1 D(C2) = −1 D(σv) = 1 D(σ v′) = −1

which is the same one-dimensional representation as that spanned by pS, and p2 spans

D(E) = 1 D(C2) = 1 D(σv) = −1 D(σ v′) = −1

which is a different one-dimensional representation; we shall denote it Γ (1)′.At this point we have found two irreducible representations of the group C2v

(Table 12.2). The two irreducible representations are normally labelled B1 and A2,respectively. An A or a B is used to denote a one-dimensional representation; A is used if the character under the principal rotation is +1, and B is used if the character is −1. Subscripts are used to distinguish the irreducible representations if there is morethan one of the same type: A1 is reserved for the representation with character 1 for all operations. When higher dimensional irreducible representations are permitted, E denotes a two-dimensional irreducible representation and T a three-dimensional irreducible representation; all the irreducible representations of C2v are one-dimensional.

There are in fact only two more species of irreducible representations of this group,for a surprising theorem of group theory states that

Number of symmetry species = number of classes (12.6)

Symmetry operations fall into the same class if they are of the same type (for example,rotations) and can be transformed into one another by a symmetry operation of the

−−

⎛⎝⎜

⎞⎠⎟

1 00 1

1 00 1−

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

1 00 1

1 00 1

⎛⎝⎜

⎞⎠⎟

�

�

�

�

A�

B�

A�

B�

Fig. 12.17 Two symmetry-adapted linearcombinations of the basis orbitals shown inFig. 12.16. The two combinations eachspan a one-dimensional irreduciblerepresentation, and their symmetry speciesare different.

Table 12.2* The C2v character table

C2v, 2mm E C2 σv σ′v h = 4

A1 1 1 1 1 z z2, y2, x2

A2 1 1 −1 −1 xy

B1 1 −1 1 −1 x zx

B2 1 −1 −1 1 y yz

* More character tables are given at the end of the Data section.


group. In C2v, for instance, there are four classes (four columns in the character table),so there are only four species of irreducible representation. The character table inTable 12.2 therefore shows the characters of all the irreducible representations of this group.

(b) The structure of character tables

In general, the columns in a character table are labelled with the symmetry operationsof the group. For instance, for the group C3v the columns are headed E, C3, and σv

(Table 12.3). The numbers multiplying each operation are the numbers of membersof each class. In the C3v character table we see that the two threefold rotations (clock-wise and counter-clockwise rotations by 120°) belong to the same class: they are related by a reflection (Fig. 12.18). The three reflections (one through each of thethree vertical mirror planes) also lie in the same class: they are related by the three-fold rotations. The two reflections of the group C2v fall into different classes: althoughthey are both reflections, one cannot be transformed into the other by any symmetryoperation of the group.

The total number of operations in a group is called the order, h, of the group. Theorder of the group C3v, for instance, is 6.

The rows under the labels for the operations summarize the symmetry propertiesof the orbitals. They are labelled with the symmetry species (the analogues of the labels σ and π). More formally, the symmetry species label the irreducible representa-tions of the group, which are the basic types of behaviour that orbitals may show whensubjected to the symmetry operations of the group, as we have illustrated for thegroup C2v. By convention, irreducible representations are labelled with upper caseRoman letters (such as A1 and E) but the orbitals to which they apply are labelled withthe lower case italic equivalents (so an orbital of symmetry species A1 is called an a1

orbital). Examples of each type of orbital are shown in Fig. 12.19.

(c) Character tables and orbital degeneracy

The character of the identity operation E tells us the degeneracy of the orbitals. Thus,in a C3v molecule, any orbital with a symmetry label a1 or a2 is nondegenerate. Anydoubly degenerate pair of orbitals in C3v must be labelled e because, in this group, onlyE symmetry species have characters greater than 1.

Because there are no characters greater than 2 in the column headed E in C3v,we know that there can be no triply degenerate orbitals in a C3v molecule. This lastpoint is a powerful result of group theory, for it means that, with a glance at the character table of a molecule, we can state the maximum possible degeneracy of its orbitals.

Table 12.3* The C3v character table

C3v, 3m E 2C3 3σv h = 6

A1 1 1 1 z z 2, x 2 + y2

A2 1 1 −1

E 2 −1 0 (x, y) (xy, x 2 − y 2), (yz, zx)

* More character tables are given at the end of the Data section.

C3C3

v

v¢v²

s

s

s

+

-

Fig. 12.18 Symmetry operations in the sameclass are related to one another by thesymmetry operations of the group. Thus,the three mirror frames shown here arerelated by threefold rotations, and the tworotations shown here are related byreflection in σv.

Comment 12.5

Note that care must be taken todistinguish the identity element E (italic,a column heading) from the symmetrylabel E (roman, a row label).


a1

a2

e

e

sN

Fig. 12.19 Typical symmetry-adapted linearcombinations of orbitals in a C3v molecule.

�

�

�

��1�1

�1

Fig. 12.20 The two orbitals shown here havedifferent properties under reflectionthrough the mirror plane: one changes sign(character −1), the other does not(character +1).

Example 12.2 Using a character table to judge degeneracy

Can a trigonal planar molecule such as BF3 have triply degenerate orbitals? What isthe minimum number of atoms from which a molecule can be built that does display triple degeneracy?

Method First, identify the point group, and then refer to the corresponding character table in the Data section. The maximum number in the column headedby the identity E is the maximum orbital degeneracy possible in a molecule of thatpoint group. For the second part, consider the shapes that can be built from two,three, etc. atoms, and decide which number can be used to form a molecule thatcan have orbitals of symmetry species T.

Answer Trigonal planar molecules belong to the point group D3h. Reference to the character table for this group shows that the maximum degeneracy is 2, as nocharacter exceeds 2 in the column headed E. Therefore, the orbitals cannot betriply degenerate. A tetrahedral molecule (symmetry group T) has an irreduciblerepresentation with a T symmetry species. The minimum number of atoms neededto build such a molecule is four (as in P4, for instance).

Self-test 12.2 A buckminsterfullerene molecule, C60 (15), belongs to the icosahedralpoint group. What is the maximum possible degree of degeneracy of its orbitals?

[5]

(d) Characters and operations

The characters in the rows labelled A and B and in the columns headed by symmetryoperations other than the identity E indicate the behaviour of an orbital under thecorresponding operations: a +1 indicates that an orbital is unchanged, and a −1 indi-cates that it changes sign. It follows that we can identify the symmetry label of the orbital by comparing the changes that occur to an orbital under each operation, andthen comparing the resulting +1 or −1 with the entries in a row of the character tablefor the point group concerned.

For the rows labelled E or T (which refer to the behaviour of sets of doubly andtriply degenerate orbitals, respectively), the characters in a row of the table are thesums of the characters summarizing the behaviour of the individual orbitals in thebasis. Thus, if one member of a doubly degenerate pair remains unchanged under asymmetry operation but the other changes sign (Fig. 12.20), then the entry is reportedas χ = 1 − 1 = 0. Care must be exercised with these characters because the transforma-tions of orbitals can be quite complicated; nevertheless, the sums of the individualcharacters are integers.

As an example, consider the O2px orbital in H2O. Because H2O belongs to the pointgroup C2v, we know by referring to the C2v character table (Table 12.2) that the labelsavailable for the orbitals are a1, a2, b1, and b2. We can decide the appropriate label forO2px by noting that under a 180° rotation (C2) the orbital changes sign (Fig. 12.21), soit must be either B1 or B2, as only these two symmetry types have character −1 underC2. The O2px orbital also changes sign under the reflection σ v′ , which identifies it asB1. As we shall see, any molecular orbital built from this atomic orbital will also be a b1 orbital. Similarly, O2py changes sign under C2 but not under σv′; therefore, it cancontribute to b2 orbitals.

The behaviour of s, p, and d orbitals on a central atom under the symmetry opera-tions of the molecule is so important that the symmetry species of these orbitals


v

v́

C2

+

-

s

s

Fig. 12.21 A px orbital on the central atom ofa C2v molecule and the symmetry elementsof the group.

sA

sB

sC

Fig. 12.22 The three H1s orbitals used toconstruct symmetry-adapted linearcombinations in a C3v molecule such asNH3.

N

�

�O

�

O�

Fig. 12.23 One symmetry-adapted linearcombination of O2px orbitals in the C2v

NO2− ion.

A B

CD

21

are generally indicated in a character table. To make these allocations, we look at the symmetry species of x, y, and z, which appear on the right-hand side of the charactertable. Thus, the position of z in Table 12.3 shows that pz (which is proportional tozf (r)), has symmetry species A1 in C3v, whereas px and py (which are proportional toxf (r) and yf (r), respectively) are jointly of E symmetry. In technical terms, we say thatpx and py jointly span an irreducible representation of symmetry species E. An s orbitalon the central atom always spans the fully symmetrical irreducible representation(typically labelled A1 but sometimes A1′) of a group as it is unchanged under all sym-metry operations.

The five d orbitals of a shell are represented by xy for dxy, etc, and are also listed onthe right of the character table. We can see at a glance that in C3v, dxy and dx 2 − y 2 on acentral atom jointly belong to E and hence form a doubly degenerate pair.

(e) The classification of linear combinations of orbitals

So far, we have dealt with the symmetry classification of individual orbitals. The sametechnique may be applied to linear combinations of orbitals on atoms that are relatedby symmetry transformations of the molecule, such as the combination ψ1 = ψA + ψB

+ ψC of the three H1s orbitals in the C3v molecule NH3 (Fig. 12.22). This combinationremains unchanged under a C3 rotation and under any of the three vertical reflectionsof the group, so its characters are

χ(E) = 1 χ(C3) = 1 χ(σv) = 1

Comparison with the C3v character table shows that ψ1 is of symmetry species A1, andtherefore that it contributes to a1 molecular orbitals in NH3.

Example 12.3 Identifying the symmetry species of orbitals

Identify the symmetry species of the orbital ψ = ψA − ψB in a C2v NO2 molecule,where ψA is an O2px orbital on one O atom and ψB that on the other O atom.

Method The negative sign in ψ indicates that the sign of ψB is opposite to that ofψA. We need to consider how the combination changes under each operation ofthe group, and then write the character as +1, −1, or 0 as specified above. Then wecompare the resulting characters with each row in the character table for the pointgroup, and hence identify the symmetry species.

Answer The combination is shown in Fig. 12.23. Under C2, ψ changes into itself,implying a character of +1. Under the reflection σv, both orbitals change sign, so ψ → −ψ , implying a character of −1. Under σ v′ , ψ → −ψ, so the character for thisoperation is also −1. The characters are therefore

χ(E) = 1 χ(C2) = 1 χ(σv) = −1 χ(σ v′) = −1

These values match the characters of the A2 symmetry species, so ψ can contributeto an a2 orbital.

Self-test 12.3 Consider PtCl 4−, in which the Cl ligands form a square planar array

of point group D4h (21). Identify the symmetry type of the combination ψA − ψB

+ ψC − ψD. [B2g]

12.5 VANISHING INTEGRALS AND ORBITAL OVERLAP 419

12.5 Vanishing integrals and orbital overlap

Suppose we had to evaluate the integral

I = �f1 f2 dτ (12.7)

where f1 and f2 are functions. For example, f1 might be an atomic orbital A on oneatom and f2 an atomic orbital B on another atom, in which case I would be their over-lap integral. If we knew that the integral is zero, we could say at once that a molecularorbital does not result from (A,B) overlap in that molecule. We shall now see thatcharacter tables provide a quick way of judging whether an integral is necessarily zero.

(a) The criteria for vanishing integrals

The key point in dealing with the integral I is that the value of any integral, and of anoverlap integral in particular, is independent of the orientation of the molecule (Fig.12.24). In group theory we express this point by saying that I is invariant under anysymmetry operation of the molecule, and that each operation brings about the trivialtransformation I → I. Because the volume element dτ is invariant under any sym-metry operation, it follows that the integral is nonzero only if the integrand itself, theproduct f1 f2, is unchanged by any symmetry operation of the molecular point group.If the integrand changed sign under a symmetry operation, the integral would be thesum of equal and opposite contributions, and hence would be zero. It follows that theonly contribution to a nonzero integral comes from functions for which under anysymmetry operation of the molecular point group f1 f2 → f1 f2, and hence for which thecharacters of the operations are all equal to +1. Therefore, for I not to be zero, theintegrand f1 f2 must have symmetry species A1 (or its equivalent in the specific mole-cular point group).

We use the following procedure to deduce the symmetry species spanned by theproduct f1 f2 and hence to see whether it does indeed span A1.

1 Decide on the symmetry species of the individual functions f1 and f2 by referenceto the character table, and write their characters in two rows in the same order as inthe table.

2 Multiply the numbers in each column, writing the results in the same order.

3 Inspect the row so produced, and see if it can be expressed as a sum of charactersfrom each column of the group. The integral must be zero if this sum does not contain A1.

For example, if f1 is the sN orbital in NH3 and f2 is the linear combination s3 = sB − sC

(Fig. 12.25), then, because sN spans A1 and s3 is a member of the basis spanning E, wewrite

f1: 1 1 1f2: 2 −1 0f1 f2: 2 −1 0

The characters 2, −1, 0 are those of E alone, so the integrand does not span A1. It followsthat the integral must be zero. Inspection of the form of the functions (see Fig. 12.25)shows why this is so: s3 has a node running through sN. Had we taken f1 = sN and f2 = s1

instead, where s1 = sA + sB + sC, then because each spans A1 with characters 1,1,1:

f1: 1 1 1f2: 1 1 1f1 f2: 1 1 1

x

y

y

x

(a)

(b)

Fig. 12.24 The value of an integral I (forexample, an area) is independent of thecoordinate system used to evaluate it. That is, I is a basis of a representation ofsymmetry species A1 (or its equivalent).

sBsC

��

Fig. 12.25 A symmetry-adapted linearcombination that belongs to the symmetryspecies E in a C3v molecule such as NH3.This combination can form a molecularorbital by overlapping with the px orbitalon the central atom (the orbital with itsaxis parallel to the width of the page; seeFig. 12.28c).


x

y

�

�

�

�

Fig. 12.26 The integral of the function f = xyover the tinted region is zero. In this case,the result is obvious by inspection, butgroup theory can be used to establishsimilar results in less obvious cases. Theinsert shows the shape of the function inthree dimensions.

x

y

�

Fig. 12.27 The integration of a function overa pentagonal region. The insert shows theshape of the function in three dimensions.

The characters of the product are those of A1 itself. Therefore, s1 and sN may havenonzero overlap. A shortcut that works when f1 and f2 are bases for irreducible repre-sentations of a group is to note their symmetry species: if they are different, then theintegral of their product must vanish; if they are the same, then the integral may benonzero.

It is important to note that group theory is specific about when an integral must bezero, but integrals that it allows to be nonzero may be zero for reasons unrelated to symmetry. For example, the N-H distance in ammonia may be so great that the(s1, sN) overlap integral is zero simply because the orbitals are so far apart.

Example 12.4 Deciding if an integral must be zero (1)

May the integral of the function f = xy be nonzero when evaluated over a region theshape of an equilateral triangle centred on the origin (Fig. 12.26)?

Method First, note that an integral over a single function f is included in the previ-ous discussion if we take f1 = f and f2 = 1 in eqn 12.7. Therefore, we need to judgewhether f alone belongs to the symmetry species A1 (or its equivalent) in the pointgroup of the system. To decide that, we identify the point group and then examinethe character table to see whether f belongs to A1 (or its equivalent).

Answer An equilateral triangle has the point-group symmetry D3h. If we refer tothe character table of the group, we see that xy is a member of a basis that spans theirreducible representation E′. Therefore, its integral must be zero, because theintegrand has no component that spans A1′.

Self-test 12.4 Can the function x2 + y2 have a nonzero integral when integratedover a regular pentagon centred on the origin? [Yes, Fig. 12.27]

In many cases, the product of functions f1 and f2 spans a sum of irreducible repres-entations. For instance, in C2v we may find the characters 2, 0, 0, −2 when we multiplythe characters of f1 and f2 together. In this case, we note that these characters are thesum of the characters for A2 and B1:

E C2v σv σ ′vA2 1 1 −1 −1B1 1 −1 1 −1A2 + B1 2 0 0 −2

To summarize this result we write the symbolic expression A2 × B1 = A2 + B1, which iscalled the decomposition of a direct product. This expression is symbolic. The × and+ signs in this expression are not ordinary multiplication and addition signs: formally,they denote technical procedures with matrices called a ‘direct product’ and a ‘directsum’. Because the sum on the right does not include a component that is a basis for anirreducible representation of symmetry species A1, we can conclude that the integralof f1 f2 over all space is zero in a C2v molecule.

Whereas the decomposition of the characters 2, 0, 0, −2 can be done by inspectionin this simple case, in other cases and more complex groups the decomposition isoften far from obvious. For example, if we found the characters 8, −2, −6, 4, it wouldnot be obvious that the sum contains A1. Group theory, however, provides a system-atic way of using the characters of the representation spanned by a product to find thesymmetry species of the irreducible representations. The recipe is as follows:

12.5 VANISHING INTEGRALS AND ORBITAL OVERLAP 421

(a)

(b)

(c)

Fig. 12.28 Orbitals of the same symmetryspecies may have non-vanishing overlap.This diagram illustrates the three bondingorbitals that may be constructed from(N2s, H1s) and (N2p, H1s) overlap in a C3v

molecule. (a) a1; (b) and (c) the twocomponents of the doubly degenerate eorbitals. (There are also three antibondingorbitals of the same species.)

1 Write down a table with columns headed by the symmetry operations of thegroup.

2 In the first row write down the characters of the symmetry species we want toanalyse.

3 In the second row, write down the characters of the irreducible representation Γwe are interested in.

4 Multiply the two rows together, add the products together, and divide by theorder of the group.

The resulting number is the number of times Γ occurs in the decomposition.

Illustration 12.1 To find whether A1 occurs in a direct product

To find whether A1 does indeed occur in the product with characters 8, −2, −6, 4 inC2v, we draw up the following table:

E C2v σv σ′v h = 4 (the order of the group)f1 f2 8 −2 −6 4 (the characters of the product)A1 1 1 1 1 (the symmetry species we are interested in)

8 −2 −6 4 (the product of the two sets of characters)

The sum of the numbers in the last line is 4; when that number is divided by theorder of the group, we get 1, so A1 occurs once in the decomposition. When theprocedure is repeated for all four symmetry species, we find that f1 f2 spans A1 + 2A2

+ 5B2.

Self-test 12.5 Does A2 occur among the symmetry species of the irreducible repre-sentations spanned by a product with characters 7, −3, −1, 5 in the group C2v?

[No]

(b) Orbitals with nonzero overlap

The rules just given let us decide which atomic orbitals may have nonzero overlap ina molecule. We have seen that sN may have nonzero overlap with s1 (the combinationsA + sB + sC), so bonding and antibonding molecular orbitals can form from (sN, s1)overlap (Fig. 12.28). The general rule is that only orbitals of the same symmetry speciesmay have nonzero overlap, so only orbitals of the same symmetry species form bond-ing and antibonding combinations. It should be recalled from Chapter 11 that theselection of atomic orbitals that had mutual nonzero overlap is the central and initialstep in the construction of molecular orbitals by the LCAO procedure. We are there-fore at the point of contact between group theory and the material introduced in thatchapter. The molecular orbitals formed from a particular set of atomic orbitals withnonzero overlap are labelled with the lowercase letter corresponding to the symmetryspecies. Thus, the (sN, s1)-overlap orbitals are called a1 orbitals (or a1*, if we wish toemphasize that they are antibonding).

The linear combinations s2 = 2sa − sb − sc and s3 = sb − sc have symmetry species E.Does the N atom have orbitals that have nonzero overlap with them (and give rise toe molecular orbitals)? Intuition (as supported by Figs. 12.28b and c) suggests thatN2px and N2py should be suitable. We can confirm this conclusion by noting that the character table shows that, in C3v, the functions x and y jointly belong to the sym-metry species E. Therefore, N2px and N2py also belong to E, so may have nonzerooverlap with s2 and s3. This conclusion can be verified by multiplying the characters


and finding that the product of characters can be expressed as the decomposition E × E = A1 + A2 + E. The two e orbitals that result are shown in Fig. 12.28 (there are also two antibonding e orbitals).

We can see the power of the method by exploring whether any d orbitals on the cen-tral atom can take part in bonding. As explained earlier, reference to the C3v charactertable shows that dz 2 has A1 symmetry and that the pairs (dx2−y2, dxy) and (dyz,dzx) eachtransform as E. It follows that molecular orbitals may be formed by (s1,dz 2) overlapand by overlap of the s2,s3 combinations with the E d orbitals. Whether or not the dorbitals are in fact important is a question group theory cannot answer because the extent of their involvement depends on energy considerations, not symmetry.

Example 12.5 Determining which orbitals can contribute to bonding

The four H1s orbitals of methane span A1 + T2. With which of the C atom orbitalscan they overlap? What bonding pattern would be possible if the C atom had dorbitals available?

Method Refer to the Td character table (in the Data section) and look for s, p, andd orbitals spanning A1 or T2.

Answer An s orbital spans A1, so it may have nonzero overlap with the A1 combina-tion of H1s orbitals. The C2p orbitals span T2, so they may have nonzero overlapwith the T2 combination. The dxy, dyz, and dzx orbitals span T2, so they may overlapthe same combination. Neither of the other two d orbitals span A1 (they span E), sothey remain nonbonding orbitals. It follows that in methane there are (C2s,H1s)-overlap a1 orbitals and (C2p,H1s)-overlap t2 orbitals. The C3d orbitals might con-tribute to the latter. The lowest energy configuration is probably a1

2t 26, with all

bonding orbitals occupied.

Self-test 12.6 Consider the octahedral SF6 molecule, with the bonding arisingfrom overlap of S orbitals and a 2p orbital on each F directed towards the central Satom. The latter span A1g + Eg + T1u. What s orbitals have nonzero overlap? Suggestwhat the ground-state configuration is likely to be.

[3s(A1g), 3p(T1u), 3d(Eg); a 21gt

61ue4

g]

(c) Symmetry-adapted linear combinations

So far, we have only asserted the forms of the linear combinations (such as s1, etc.) thathave a particular symmetry. Group theory also provides machinery that takes an arbitrary basis, or set of atomic orbitals (sA, etc.), as input and generates combinationsof the specified symmetry. Because these combinations are adapted to the symmetryof the molecule, they are called symmetry-adapted linear combinations (SALC).Symmetry-adapted linear combinations are the building blocks of LCAO molecularorbitals, for they include combinations such as those used to construct molecular orbitals in benzene. The construction of SALCs is the first step in any molecular orbitaltreatment of molecules.

The technique for building SALCs is derived by using the full power of group theory. We shall not show the derivation (see Further reading), which is very lengthy,but present the main conclusions as a set of rules:

1 Construct a table showing the effect of each operation on each orbital of the original basis.

12.6 VANISHING INTEGRALS AND SELECTION RULES 423

2 To generate the combination of a specified symmetry species, take each columnin turn and:

(i) Multiply each member of the column by the character of the correspondingoperation.

(ii) Add together all the orbitals in each column with the factors as determinedin (i).

(iii) Divide the sum by the order of the group.

For example, from the (sN,sA,sB,sC) basis in NH3 we form the table shown in the margin. To generate the A1 combination, we take the characters for A1 (1,1,1,1,1,1);then rules (i) and (ii) lead to

ψ ∝ sN + sN + · · · = 6sN

The order of the group (the number of elements) is 6, so the combination of A1 sym-metry that can be generated from sN is sN itself. Applying the same technique to thecolumn under sA gives

ψ = 1–6(sA + sB + sC + sA + sB + sC ) = 1–3 (sA + sB + sC)

The same combination is built from the other two columns, so they give no further information. The combination we have just formed is the s1 combination we used before (apart from the numerical factor).

We now form the overall molecular orbital by forming a linear combination of allthe SALCs of the specified symmetry species. In this case, therefore, the a1 molecularorbital is

ψ = cNsN + c1s1

This is as far as group theory can take us. The coefficients are found by solving theSchrödinger equation; they do not come directly from the symmetry of the system.

We run into a problem when we try to generate an SALC of symmetry species E, because, for representations of dimension 2 or more, the rules generate sums ofSALCs. This problem can be illustrated as follows. In C3v, the E characters are 2, −1,−1, 0, 0, 0, so the column under sN gives

ψ = 1–6 (2sN − sN − sN + 0 + 0 + 0 ) = 0

The other columns give1–6(2sA − sB − sC) 1–6 (2sB − sA − sC) 1–6(2sC − sB − sA)

However, any one of these three expressions can be expressed as a sum of the othertwo (they are not ‘linearly independent’). The difference of the second and third gives1–2(sB − sC), and this combination and the first, 1–6(2sA − sB − sC) are the two (now linearlyindependent) SALCs we have used in the discussion of e orbitals.

12.6 Vanishing integrals and selection rules

Integrals of the form

I = �f1 f2 f3dτ (12.8)

are also common in quantum mechanics for they include matrix elements of oper-ators (Section 8.5d), and it is important to know when they are necessarily zero. Forthe integral to be nonzero, the product f1 f2 f3 must span A1 (or its equivalent) or containa component that spans A1. To test whether this is so, the characters of all three func-tions are multiplied together in the same way as in the rules set out above.

Original basis

sN sA sB sC

Under E sN sA sB sC

C 3+ sN sB sC sA

C3− sN sC sA sB

σv sN sA sC sB

σ v′ sN sB sA sC

σ v″ sN sC sB sA


Forbidden

Forbidden

x-po

lari

zed

x-po

lari

zed

A1 A2

B1 B2

y-polarized

Fig. 12.29 The polarizations of the allowedtransitions in a C2v molecule. The shadingindicates the structure of the orbitals of thespecified symmetry species. Theperspective view of the molecule makes itlook rather like a door stop; however, fromthe side, each ‘door stop’ is in fact anisosceles triangle.

Example 12.6 Deciding if an integral must be zero (2)

Does the integral ∫(3dz2)x(3dxy)dτ vanish in a C2v molecule?

Method We must refer to the C2v character table (Table 12.2) and the characters of the irreducible representations spanned by 3z2 − r2 (the form of the dz2 orbital),x, and xy; then we can use the procedure set out above (with one more row of multiplication).

Answer We draw up the following table:

E C2 σv σ ′vf3 = dxy 1 1 −1 −1 A2

f2 = x 1 −1 1 −1 B1

f1 = dz2 1 1 1 1 A1

f1 f2 f3 1 −1 −1 1

The characters are those of B2. Therefore, the integral is necessarily zero.

Self-test 12.7 Does the integral ∫(2px)(2py)(2pz)dτ necessarily vanish in an octa-hedral environment? [No]

We saw in Chapters 9 and 10, and will see in more detail in Chapters 13 and 14, thatthe intensity of a spectral line arising from a molecular transition between some initialstate with wavefunction ψi and a final state with wavefunction ψf depends on the(electric) transition dipole moment, µfi. The z-component of this vector is definedthrough

µz,fi = −e�ψ f*zψi dτ [12.9]

where −e is the charge of the electron. The transition moment has the form of the integral in eqn 12.8, so, once we know the symmetry species of the states, we can usegroup theory to formulate the selection rules for the transitions.

As an example, we investigate whether an electron in an a1 orbital in H2O (whichbelongs to the group C2v) can make an electric dipole transition to a b1 orbital(Fig. 12.29). We must examine all three components of the transition dipole moment,and take f2 in eqn 12.8 as x, y, and z in turn. Reference to the C2v character table showsthat these components transform as B1, B2, and A1, respectively. The three calcula-tions run as follows:

x-component y-component z-component

E C2 σv σ′v E C2 σv σ′v E C2 sv s′v

f3 1 −1 1 −1 1 −1 1 −1 1 −1 1 −1B1

f2 1 −1 1 −1 1 −1 −1 1 1 1 1 1

f1 1 1 1 1 1 1 1 1 1 1 1 1A1

f1 f2 f3 1 1 1 1 1 1 −1 −1 1 −1 1 −1

Only the first product (with f2 = x) spans A1, so only the x-component of the tran-sition dipole moment may be nonzero. Therefore, we conclude that the electric dipole

CHECKLIST OF KEY IDEAS 425

Checklist of key ideas

1. A symmetry operation is an action that leaves an objectlooking the same after it has been carried out.

2. A symmetry element is a point, line, or plane with respect towhich a symmetry operation is performed.

3. A point group is a group of symmetry operations that leaves atleast one common point unchanged. A space group, a groupof symmetry operations that includes translation throughspace.

4. The notation for point groups commonly used for moleculesand solids is summarized in Table 12.1.

5. To be polar, a molecule must belong to Cn, Cnv, or Cs (andhave no higher symmetry).

6. A molecule may be chiral only if it does not possess an axis ofimproper rotation, Sn.

7. A representative D(X) is a matrix that brings about thetransformation of the basis under the operation X. The basis isthe set of functions on which the representative acts.

8. A character, χ, is the sum of the diagonal elements of a matrixrepresentative.

9. A character table characterizes the different symmetry typespossible in the point group.

10. In a reduced representation all the matrices have block-diagonal form. An irreducible representation cannot bereduced further.

11. Symmetry species are the labels for the irreduciblerepresentations of a group.

12. Decomposition of the direct product is the reduction of aproduct of symmetry species to a sum of symmetry species, Γ × Γ ′ = Γ (1) + Γ (2) + · · ·

transitions between a1 and b1 are allowed. We can go on to state that the radiationemitted (or absorbed) is x-polarized and has its electric field vector in the x-direction,because that form of radiation couples with the x–component of a transition dipole.

Example 12.7 Deducing a selection rule

Is px → py an allowed transition in a tetrahedral environment?

Method We must decide whether the product py qpx , with q = x, y, or z, spans A1 byusing the Td character table.

Answer The procedure works out as follows:

E 8C3 3C2 6σd 6S4

f3(py) 3 0 −1 −1 1 T2

f2(q) 3 0 −1 −1 1 T2

f1(px) 3 0 −1 −1 1 T2

f1 f2 f3 27 0 −1 −1 1

We can use the decomposition procedure described in Section 12.5a to deduce thatA1 occurs (once) in this set of characters, so px → py is allowed.

A more detailed analysis (using the matrix representatives rather than the char-acters) shows that only q = z gives an integral that may be nonzero, so the transitionis z-polarized. That is, the electromagnetic radiation involved in the transition hasits electric vector aligned in the z-direction.

Self-test 12.8 What are the allowed transitions, and their polarizations, of a b1

electron in a C4v molecule? [b1 → b1(z); b1 → e(x,y)]

The following chapters will show many more examples of the systematic use ofsymmetry. We shall see that the techniques of group theory greatly simplify the ana-lysis of molecular structure and spectra.

atkins' physical chemistry - ohio northernj-gray/chem3421/atkins-symmetry.pdf · molecular...

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